{ "data": { "question": { "questionId": "2255", "questionFrontendId": "2134", "categoryTitle": "Algorithms", "boundTopicId": 1196321, "title": "Minimum Swaps to Group All 1's Together II", "titleSlug": "minimum-swaps-to-group-all-1s-together-ii", "content": "

A swap is defined as taking two distinct positions in an array and swapping the values in them.

\n\n

A circular array is defined as an array where we consider the first element and the last element to be adjacent.

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Given a binary circular array nums, return the minimum number of swaps required to group all 1's present in the array together at any location.

\n\n

 

\n

Example 1:

\n\n
\nInput: nums = [0,1,0,1,1,0,0]\nOutput: 1\nExplanation: Here are a few of the ways to group all the 1's together:\n[0,0,1,1,1,0,0] using 1 swap.\n[0,1,1,1,0,0,0] using 1 swap.\n[1,1,0,0,0,0,1] using 2 swaps (using the circular property of the array).\nThere is no way to group all 1's together with 0 swaps.\nThus, the minimum number of swaps required is 1.\n
\n\n

Example 2:

\n\n
\nInput: nums = [0,1,1,1,0,0,1,1,0]\nOutput: 2\nExplanation: Here are a few of the ways to group all the 1's together:\n[1,1,1,0,0,0,0,1,1] using 2 swaps (using the circular property of the array).\n[1,1,1,1,1,0,0,0,0] using 2 swaps.\nThere is no way to group all 1's together with 0 or 1 swaps.\nThus, the minimum number of swaps required is 2.\n
\n\n

Example 3:

\n\n
\nInput: nums = [1,1,0,0,1]\nOutput: 0\nExplanation: All the 1's are already grouped together due to the circular property of the array.\nThus, the minimum number of swaps required is 0.\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "最少交换次数来组合所有的 1 II", "translatedContent": "

交换 定义为选中一个数组中的两个 互不相同 的位置并交换二者的值。

\n\n

环形 数组是一个数组,可以认为 第一个 元素和 最后一个 元素 相邻

\n\n

给你一个 二进制环形 数组 nums ,返回在 任意位置 将数组中的所有 1 聚集在一起需要的最少交换次数。

\n\n

 

\n\n

示例 1:

\n\n
输入:nums = [0,1,0,1,1,0,0]\n输出:1\n解释:这里列出一些能够将所有 1 聚集在一起的方案:\n[0,0,1,1,1,0,0] 交换 1 次。\n[0,1,1,1,0,0,0] 交换 1 次。\n[1,1,0,0,0,0,1] 交换 2 次(利用数组的环形特性)。\n无法在交换 0 次的情况下将数组中的所有 1 聚集在一起。\n因此,需要的最少交换次数为 1 。\n
\n\n

示例 2:

\n\n
输入:nums = [0,1,1,1,0,0,1,1,0]\n输出:2\n解释:这里列出一些能够将所有 1 聚集在一起的方案:\n[1,1,1,0,0,0,0,1,1] 交换 2 次(利用数组的环形特性)。\n[1,1,1,1,1,0,0,0,0] 交换 2 次。\n无法在交换 0 次或 1 次的情况下将数组中的所有 1 聚集在一起。\n因此,需要的最少交换次数为 2 。\n
\n\n

示例 3:

\n\n
输入:nums = [1,1,0,0,1]\n输出:0\n解释:得益于数组的环形特性,所有的 1 已经聚集在一起。\n因此,需要的最少交换次数为 0 。
\n\n

 

\n\n

提示:

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